Cellular covers of divisible uniserial modules over valuation domains (Q6612611)

Let \(R\) be an integral domain with field of fractions \(Q\). A morphism \(\gamma\colon \Gamma \to M\) in \({\mathrm{Mod}\text{-}R}\) is a \textit{cellular cover} if it induces an isomorphism \(\Hom_R(\Gamma,\Gamma)\overset\cong\longrightarrow \Hom_R(\Gamma,M)\). Moreover, the following short exact sequence in \(\mathrm{Mod}\text{-}R\) \N\[\N0\to K:=\mathrm{Ker}(\gamma)\longrightarrow \Gamma \overset{\gamma}\longrightarrow M\to 0 \tag{$*$}\N\] \Nis said to be \textit{cellular} if \(\gamma\) is a cellular cover. Given a module \(M\), it is natural to look for some classification of its (surjective) cellular covers, either in general or requiring that \(\Gamma\) (and/or \(K\)) belong to suitable classes of modules. We refer to the well-written introduction of the paper for a detailed discussion of the state of the art on this kind of problems; the best known results are for cellular sequences of Abelian groups, when \(M\) is divisible.\N\NThe goal of the paper under review is to(at least partially) extend the results for cellular covers of divisible Abelian groups, to general (\(h\)-)divisible \(R\)-modules. The main results in this generality are the following (see Section~3 of the paper):\N\begin{itemize}\N\item[--] for \(M\) torsion-free and divisible, \((*)\) is cellular only if \(K=0\);\N\item[--] for \(M\) torsion and \(h\)-divisible, \((*)\) is cellular if and only if it is isomorphic to \(0\to \Hom_R(Q/R,M)\to \Hom_R(Q,M)\to M\to 0\), relating \(M\) with its Matlis dual.\N\end{itemize}\N\NStarting from Section~4, \(R\) is assumed to be a valuation domain, and \(M\) a torsion, divisible and uniserial \(R\)-module. In particular, in Sections~4 and 5, \(M\) (and, for the best results, also \(\Gamma\)) is assumed to be standard uniserial while, in Section~6, \(M\) (and, for the best results, also \(K\)) is assumed to be a ``clone of \(Q/R\)'' (these are a class of examples of uniserial divisible modules that are non-standard and, as a consequence, not \(h\)-divisible).